How do I integrate $x^5(4+x^2)^{-1/2}$?

Question

I just started learning substitution and I can't seem to solve this exercise. I'm using $x = 2sinh(t)$

Full Solution:

$32\int(sinh(t)*(cosh^2(t)-1)^2=$

$32\int(sinh(t)*(cosh^4(t)-2cosh^2(t)+1)=$

$32\int(cosh^4(t)sinh(t)-2cosh^2(t)sinh(t)+sinh(t))=$

$32(\int(cosh^4(arcsinx)sinh(arcsinx)-2\int(cosh^2(arcsinx)sinh(arcsinhx)+\int(sinh(arcsinx)))=$

$32(\int((1-x^2)^2*x) - 2\int((1-x^2)*x) + \int(x))=$

Answer 1

Hint: Let $4+x^2=u^2$ and note that $x^5=(x)(x^4)=(x)(u^2-4)^2$.

But if you want to use a $\sinh$ substitution, let $x=2\sinh t$. You will end up integrating a constant times $\sinh^5 t$, which you can rewrite as $(\sinh t)(\cosh^2 t-1)^2$. And now another substitution.

Added detail Let $4+x^2=u^2$, with $u$ positive. Then $2x\,dx=2u\,du$ and therefore $x\,dx=u\,du$. Note that $(4+x^2)^{-1/2}=\frac{1}{u}$. We have $x^5=(x)(x^4)=x(u^2-4)^2$. Substituting we find that our integral is $$\int \frac{(u^2-4)^2 u}{u}\,du.$$ Thus we want $$\int (u^4-8u^2+16)\,du.$$ This is equal to $\frac{u^5}{5}-\frac{8u^3}{3}+16u+C$.

Finally, replace $u$ by $(4+x^2)^{1/2}$.

Answer 2

Notice, let, $x=2\tan \theta\ \implies dx=2\sec^2\theta \ d\theta $ $$\int \frac{x^5}{\sqrt{4+x^2}}\ dx$$$$=\int \frac{(2\tan\theta)^5}{\sqrt{4+4\tan^2\theta}}(2\sec^2\theta \ d\theta)$$ $$=32\int \frac{\tan^5\theta \sec^2\theta}{|\sec\theta|}\ d\theta$$ Assuming $0<\theta<\pi/2$ $$=32\int \frac{\tan^5\theta \sec^2\theta}{\sec\theta}\ d\theta$$ $$=32\int (\tan^2\theta)^2 \sec\theta\tan\theta\ d\theta$$

$$=32\int (\sec^2\theta-1)^2 \sec\theta\tan\theta\ d\theta$$ $$=32\int (\sec^4\theta-2\sec^2\theta+1) \ d(\sec\theta)$$ $$=32\left(\frac{\sec^5\theta}{5}-\frac{2\sec^3\theta}{3}+\sec\theta\right)+C$$

Answer 3

$$\int \frac{x^5}{(4+x^2)^{1/2}} dx$$

It seems like you're trying to take advantage of the identity $1+\sinh^2 x = \cosh^2 x$. But note that the term in parenthesis is not quite of this form. We need to accommodate for the $4$; thus we want to get $4+4\sinh^2 u$. So let $x = 2\sinh u$ and $dx = 2\cosh u du$. Then,

$$\int \frac{2^5\sinh^5 u}{(4+4\sinh^2 u)^{1/2}} 2\cosh u\ du$$

$$= 32\int \frac{\sinh^5 u}{2(1+\sinh^2 u)^{1/2}} 2\cosh u\ du$$

$$= 32 \int \frac{\sinh^5 u}{2\cosh u} 2\cosh u\ du$$

$$= 32 \int \sinh^5 u\ du$$

From here you can use a reduction formula or apply integration by parts.